PHP:foreach(额外数据显示)

时间:2016-09-05 05:33:31

标签: php foreach

额外数据显示:

http://image.prntscr.com/image/41c1fbbdc1354ed2801f5afcfc5b4e0d.png

我从foreach循环中获得了一些奇怪的数据。我的日子和时间重复两次。 (请参阅附图)。但我不想在最后一行显示数据。可以请别人帮忙解决方案。感谢

我的代码在这里: -

<button class="btn btn-primary" type="button" data-toggle="collapse" data-target="#collapseExample" aria-expanded="false" aria-controls="collapseExample"> Timings </button>
<div class="collapse" id="collapseExample">
<div class="well">
<?php

      date_default_timezone_set('Australia/Melbourne');
      $datetime = new \DateTime();

      $listItem = array('<li class="active">', '</li>');

     $curDay = date('l');

      $status = array( "Now Open" ,"Closed",  "Opening Soon",  "Closing Soon", "Open 24 hours", "  ");

      $times = $times = array(
        1 => array('day' => 'Monday', 'open' => date( $row_DetailRS1['monO'] ), 'close' => date( $row_DetailRS1['monC'] )),
        2 => array('day' => 'Tuesday', 'open' => date( $row_DetailRS1['tueO'] ), 'close' => date( $row_DetailRS1['tueC'] )),
        3 => array('day' => 'Wednesday', 'open' => date( $row_DetailRS1['wedO'] ), 'close' => date( $row_DetailRS1['wedC']) ),
        4 => array('day' => 'Thursday', 'open' => date( $row_DetailRS1['thurO'] ), 'close' => date( $row_DetailRS1['thurC']) ),
        5 => array('day' => 'Friday', 'open' => date( $row_DetailRS1['friO'] ), 'close' => date( $row_DetailRS1['friC'] )),
        6 => array('day' => 'Saturday', 'open' => date( $row_DetailRS1['satO'] ), 'close' => date( $row_DetailRS1['satC']) ),
        7 => array('day' => 'Sunday', 'open' => date( $row_DetailRS1['sunO'] ), 'close' => date( $row_DetailRS1['sunC'] )) );



      $html .= "  <a href=''>"; echo $curDay; $html .="</a>

                   <table class='table table-striped'  border='0' align='center' cellpadding='10' cellspacing='20'>
                   <tr>
                     <td>Days</td>
                     <td><span class='white-text' style='margin-right: 3em;'></td> 
                     <td>Business Hours</td> 
                     <td><span class='white-text' style='margin-right: 3em;'></td>
                     <td>Hours</td>
                    </tr>";
      $i = 1;



    $cd = $datetime->format('N'); 

    $timenow =     date("H:i:s", time());


    // Create an array of day numbers that start with current day and loops around
    $day_order = array_merge(range($cd, 7), range(1, $cd-1));



     foreach ($day_order as $daynum): {
        $oc = $times[$daynum];
        $openingTime = $oc['open'];
        $closingTime = $oc['close'];
        $openingSoon = date('H:i:sA', strtotime($openingTime)-3600);
        $closingSoon = date('H:i:sA', strtotime($closingTime)-3600);
        if ($cd == $daynum) {
            if ($openingTime == ' ' && $closingTime == ' ') {
                $s = $status[4];
            }
            elseif ($timenow < $openingSoon || $timenow > $closingTime ) {
                $s = $status[1];
            }
            elseif  ($timenow > $openingSoon && $timenow < $openingTime ) {
                $s = $status[2];
            }
            elseif  ($timenow > $closingSoon && $timenow < $closingTime ) {
                $s = $status[3];
            } else {
                $s = $status[0];
            }
        } else {
            $s = " ";
        }

        $html .= "<tr>";
        $html .= "<td>". $oc['day']."</td>";
        $html .= "<td>  <span class='white-text' style='margin-right: 3em;'> </td>";
        $html .= "<td>".$openingTime." to ".$closingTime."</td>";
        $html .= "<td>  <span class='white-text' style='margin-right: 3em;'> </td>";
        $html .= "<td>".$s."</td>";
        $html .= "</tr>";

    }
          $datetime->add(new \DateInterval('P1D'));








      endforeach;
      $html .= "</table>";
      echo $html;



     ?>

2 个答案:

答案 0 :(得分:2)

$day_order = array_merge(range($cd, 7), range(1, $cd-1));

如果var_dump $cd == 1,那么你将获得

[1,2,3,4,5,6,7,1,0];

你需要这样做:

$day_order = range($cd, 7);
if ($cd != 1) {
    $day_order = array_merge($day_order, range(1, $cd-1));
}

答案 1 :(得分:1)

我认为错误就在于您定义foreach范围的行:

$day_order = array_merge(range($cd, 7), range(1, $cd-1));

必须是:

if ($cd != 1) {
    $day_order = array_merge(range($cd, 7), range(1, $cd-1));
} else {
    $day_order = range($cd, 7));
}

原因是因为当$cd == 1时它会添加一个&#34; 0&#34;到数字范围,因为range(1,0)是&#34; 1,0&#34;。您必须通过检查$cd == 1

来拦截此添加